Rewrite the function using angle identities
Given
f ( x ) = sin x + cos x − 2 sin x − cos x . f(x)=\frac{\sin x+\cos x-\sqrt2}{\sin x-\cos x}. f ( x ) = sin x − cos x sin x + cos x − 2 . Use
sin x + cos x = 2 sin ( x + π 4 ) , sin x − cos x = 2 sin ( x − π 4 ) . \sin x+\cos x=\sqrt2\sin\left(x+\frac\pi4\right),
\qquad
\sin x-\cos x=\sqrt2\sin\left(x-\frac\pi4\right). sin x + cos x = 2 sin ( x + 4 π ) , sin x − cos x = 2 sin ( x − 4 π ) . So
f ( x ) = 2 sin ( x + π 4 ) − 2 2 sin ( x − π 4 ) = sin ( x + π 4 ) − 1 sin ( x − π 4 ) . f(x)=\frac{\sqrt2\sin\left(x+\frac\pi4\right)-\sqrt2}{\sqrt2\sin\left(x-\frac\pi4\right)}
=\frac{\sin\left(x+\frac\pi4\right)-1}{\sin\left(x-\frac\pi4\right)}. f ( x ) = 2 sin ( x − 4 π ) 2 sin ( x + 4 π ) − 2 = sin ( x − 4 π ) sin ( x + 4 π ) − 1 . Let
y = x − π 4 . y=x-\frac\pi4. y = x − 4 π . Then
x + π 4 = y + π 2 , x+\frac\pi4=y+\frac\pi2, x + 4 π = y + 2 π , so
sin ( x + π 4 ) = sin ( y + π 2 ) = cos y . \sin\left(x+\frac\pi4\right)=\sin\left(y+\frac\pi2\right)=\cos y. sin ( x + 4 π ) = sin ( y + 2 π ) = cos y . Hence
f ( x ) = cos y − 1 sin y . f(x)=\frac{\cos y-1}{\sin y}. f ( x ) = sin y cos y − 1 . Now simplify:
cos y − 1 sin y = − 2 sin 2 ( y / 2 ) 2 sin ( y / 2 ) cos ( y / 2 ) = − tan y 2 . \frac{\cos y-1}{\sin y}
=\frac{-2\sin^2(y/2)}{2\sin(y/2)\cos(y/2)}
=-\tan\frac y2. sin y cos y − 1 = 2 sin ( y /2 ) cos ( y /2 ) − 2 sin 2 ( y /2 ) = − tan 2 y . Therefore
f ( x ) = − tan ( x − π / 4 2 ) = − tan ( x 2 − π 8 ) . f(x)=-\tan\left(\frac{x-\pi/4}{2}\right)= -\tan\left(\frac x2-\frac\pi8\right). f ( x ) = − tan ( 2 x − π /4 ) = − tan ( 2 x − 8 π ) .
Differentiate twice
Let
u = x 2 − π 8 . u=\frac x2-\frac\pi8. u = 2 x − 8 π . Then
f ( x ) = − tan u . f(x)=-\tan u. f ( x ) = − tan u . So
f ′ ( x ) = − sec 2 u ⋅ 1 2 = − 1 2 sec 2 u . f'(x)=-\sec^2 u\cdot \frac12=-\frac12\sec^2 u. f ′ ( x ) = − sec 2 u ⋅ 2 1 = − 2 1 sec 2 u . Differentiate again:
f ′ ′ ( x ) = − 1 2 ⋅ d d x ( sec 2 u ) . f''(x)=-\frac12\cdot \frac{d}{dx}(\sec^2 u). f ′′ ( x ) = − 2 1 ⋅ d x d ( sec 2 u ) . Now
d d x ( sec 2 u ) = 2 sec 2 u tan u ⋅ 1 2 = sec 2 u tan u . \frac{d}{dx}(\sec^2 u)=2\sec^2 u\tan u\cdot \frac12=\sec^2 u\tan u. d x d ( sec 2 u ) = 2 sec 2 u tan u ⋅ 2 1 = sec 2 u tan u . Thus
f ′ ′ ( x ) = − 1 2 sec 2 u tan u . f''(x)=-\frac12\sec^2 u\tan u. f ′′ ( x ) = − 2 1 sec 2 u tan u .
Evaluate at x = 7 π 12 x=\frac{7\pi}{12} x = 12 7 π
First compute u u u
u = 1 2 ⋅ 7 π 12 − π 8 = 7 π 24 − 3 π 24 = π 6 . u=\frac12\cdot\frac{7\pi}{12}-\frac\pi8
=\frac{7\pi}{24}-\frac{3\pi}{24}
=\frac\pi6. u = 2 1 ⋅ 12 7 π − 8 π = 24 7 π − 24 3 π = 6 π . Hence
f ( 7 π 12 ) = − tan π 6 = − 1 3 . f\left(\frac{7\pi}{12}\right)=-\tan\frac\pi6=-\frac1{\sqrt3}. f ( 12 7 π ) = − tan 6 π = − 3 1 . Also
f ′ ′ ( 7 π 12 ) = − 1 2 sec 2 π 6 tan π 6 . f''\left(\frac{7\pi}{12}\right)=-\frac12\sec^2\frac\pi6\tan\frac\pi6. f ′′ ( 12 7 π ) = − 2 1 sec 2 6 π tan 6 π . Since
sec 2 π 6 = 1 cos 2 ( π / 6 ) = 1 ( 3 / 2 ) 2 = 4 3 , tan π 6 = 1 3 , \sec^2\frac\pi6=\frac{1}{\cos^2(\pi/6)}=\frac{1}{(\sqrt3/2)^2}=\frac43,
\qquad
\tan\frac\pi6=\frac1{\sqrt3}, sec 2 6 π = cos 2 ( π /6 ) 1 = ( 3 /2 ) 2 1 = 3 4 , tan 6 π = 3 1 , we get
f ′ ′ ( 7 π 12 ) = − 1 2 ⋅ 4 3 ⋅ 1 3 = − 2 3 3 . f''\left(\frac{7\pi}{12}\right)=-\frac12\cdot \frac43\cdot \frac1{\sqrt3}
=-\frac{2}{3\sqrt3}. f ′′ ( 12 7 π ) = − 2 1 ⋅ 3 4 ⋅ 3 1 = − 3 3 2 .
Compute the product
f ( 7 π 12 ) f ′ ′ ( 7 π 12 ) = ( − 1 3 ) ( − 2 3 3 ) = 2 9 . f\left(\frac{7\pi}{12}\right)f''\left(\frac{7\pi}{12}\right)
=\left(-\frac1{\sqrt3}\right)\left(-\frac{2}{3\sqrt3}\right)
=\frac{2}{9}. f ( 12 7 π ) f ′′ ( 12 7 π ) = ( − 3 1 ) ( − 3 3 2 ) = 9 2 .
Check options
The value is
2 9 . \boxed{\frac{2}{9}}. 9 2 . So the correct option is B .